Name: Multiple choice questions. Pick the BEST answer (2 pts ea)

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1 Exam Oct. 5, 1999 Multiple choice questions. Pick the BEST answer (2 pts ea) 1. The lipids of a red blood cell membrane are all a. phospholipids b. amphipathic c. glycolipids d. unsaturated 2. The fluidity of membrane lipids is: a. higher at cold than warm temperatures b. increased by the presence of longer fatty acids c. increased by the presence of saturated fatty acids d. limited to the plane of the membrane 3. The composition of phospholipids can be described as: a. phosphatidic acid esterified with an alcohol head group. b. diacylglycerol and unsaturated fatty acids c. sphingosine d. diacylglycerol esterified with an alcohol head group. 4. Extrinsic proteins are a. proteins on the outside of the cell b. proteins that require a detergent to extract c. proteins that can be extracted by varying the salt concentration d. proteins that can be cleaved by trypsin. 5. In the red blood cell membrane the membrane cytoskeleton contains a. actin filaments bound to spectrin monomers bound to band 3. b. actin filaments bound to spectrin tetramers bound to ankyrin c. band 3 bound to glycophorin via ankyrin. d. glycophorin bound to spectrin. 6. In the plasma membrane a. all of the phosphatidylcholine and carbohydrates are on the outside of the cell. b. all of the sugars are on the outside and all of the disulfide bonds are on the inside. c. proteins on the outside can be iodinated by lactoperoxidase added to the outside. d. all of the "GPI-linked" proteins are on the inside of the cell. 7. FRAP can be used to: a. identify membrane proteins b. determine if the protein is on the outside or inside of the membrane c. determine if the protein is free to move in the plane of the membrane d. determine if the protein is on the basolateral surface. 8. The signal recognition particle (SRP) a. binds to a nascent signal sequence and stops further synthesis b. recognizes the mrna coding for a hydrophobic stretch of amino acids c. recognizes the translocators in the RER d. is needed to bring each start transfer sequence to the membrane. 1

2 9. The functions of the RER include: a. synthesis of phospholipids, synthesis of core N-linked oligosaccharides. b. folding of proteins and synthesis of O-linked oligosaccharides. c. formation of disulfide bonds and addition of terminal sialic acids. d. removal of the signal peptide and phosphorylation of mannose groups 10.A protein that crosses the membrane 7 times has: a. 7 hydrophobic stretches of 15 to 20 amino acids b. up to 7 hydrophobic stretches of 15 to 20 amino acids c. it amino and carboxy-termini on the same side of the membrane d. 7 or 8 hydrophobic stretches of 15 to 20 amino acids. 11. The movement of a protein on an SDS gel is best described as proportional to : a. the negativecharge b. the molecular weight c. the log of the carbohydrate content d. the log of the molecular weight 12. Viruses are not cells because: a. they are too small b. they often have RNA rather than DNA as their genetic material c. they are not capable of reproduction on their own. d. they don't have a proper plasma membrane. 13. Dolichol a. is a fatty acid present in the plasma membrane b. is a long polyisoprenoid present in the Golgi c. is required in the synthesis of N-linked oligosaccharide core d. is a small GTPase 14. The "default" pathway from the trans Golgi refers to a. transfer of normal lysosomal enzymes to lysosomes b. the pathway mediated by clathrin coated vesicles. c. the pathway used by constitutive secretion. d. the pathway returning Golgi enzymes to the Cis and Medial Golgi stacks 15. You have made a chimeric protein between green fluorescent protein and a protein of teh COPII complex. When viewed in the fluorescent microscope, you expect to see fluorescence moving between: a. the Golgi and the nucleus b. the Golgi and the lysosomes c. the plasma membrane and mitochondria d. RER and Golgi. 2

3 16. Consider a protein that contains an ER signal sequence at its amino terminus and a nuclear localization sequence in its middle. What do you think the fate of this protein would be? Explain your answer (5 pts) The protein will either be an integral membrane protein or, if the signal sequence is cleaved it will be secreted. The protein will be translocated co-translationally so the nuclear localization sequence will never be available in the cytoplasm --so it can not be transported to the nucleus. 17. Write the structure for the following peptide SCIQKD at ph 7. What is its net charge at ph 11.5? At ph 2.5? (10 pts) At ph 11.5, the two amino groups are not protonated, the cysteine and the two carboxyl groups are fully ionized, so the net charge is 3 At ph 2.5, the amino groups and cysteine are fully protonated, but the carboxy terminal carboxyl group is half protonated or has a net charge of about 0.5. The carboxyl group of the aspartate (pk=3.9) is more than 90% protonated and hence bears a net charge of < The total net charge is in the range of +2 + (-.06) =

4 18. Compare and contrast the properties of an α-helix and β-sheet. Include in your discussion the different types of β-sheet that are present in proteins. Feel free to use diagrams. (13 pts) Similarities: In both cases the hydrogen bonds are as linear as possible, although the beta-sheet has an intrinsic variation along the path of the polypeptide due to twisting of the backbone. Both structures also maximize the number of hydrogen bonds between the carbonyl oxygen and the amide hydrogen of the peptide bond. Differences: The alpha-helix has a handedness, it is right-handed with a pitch of 0.54 nm and a rise of 0.15 nm (the major and minor periodicities). This results in 3.6 amino acid residues per turn of the helix. The helix also permits the accommodation of large residues, since the resides are spun out from the peptide backbone. The beta-sheet has a repeat of about 0.7 nm that is attributed to the side chains. Because the side chains of alternating amino acids lie in the same plane, small R groups are favored. Within a protein the beta-helix can form both parallel and anti-parallel conformations. When parallel they can assume a beta-barrel or saddle motif. The parallel structures have a right-handed structure in that the connecting polypeptide lies above the plane if the direction of the polypeptide backbone is from south to north. The beta-sheet can also adopt a structure in which sheets are composed of anti-parallel structures. In this case, they exist as either pairs of apposed sheets, or a sheet covered by a alphahelix/random coil. For anti-parallel, the turn that is intrinsic to the sheet usually contains a proline and glycine if the turn is tight. 4

5 19. Discuss the experimental evidence that primary structure dictates tertiary structure. (15 pts) RNase has 4 specific disulfide linkages. These linkages form when the protein has the correct conformation. The protein in the correct conformation is enzymatically active. When RNase is denatured under reducing conditions that break the disulfide bonds, the protein is no longer enzymatically active and assumes a random coil structure. Upon renaturation under oxidizing conditions that permit the formation of disulfide bonds, one observes that 95% of the enzymatic activity is restored and that in 95% of the instances that the correct disulfide bonds form, indicating that the renatured protein has assumed the correct conformation. If the process were random, then only ~1% of the activity would have been restored and only ~1% of the initial enzyme activity restored. 5

6 20. You have isolated two proteins. Protein A is a homotetramer. Each subunit contains 525 amino acids and the subunit is spherical. Protein B is a monomer that also consists of 525 amino acids and the subunit is also spherical. The molar fraction of hydrophobic amino acid of protein A should be greater than, equal to, or less than the molar fraction of hydrophobic amino acids in protein B? State your line of reasoning. (6 pts) The molar fraction of hydrophobic amino acids in protein A should be larger than that in protein B. The reason for this is that hydrophobic amino acids tend to reside in the interior of the protein since they cannot interact with the water solvent by readily establishing hydrogen bonds that would permit them to intercalate into the water structure. For the subunits to interact in an aqueous environment, the surface of the interacting subunits is usually enriched with hydrophobic amino acids that then interact with other. This results in interacting surfaces in which the hydrophobic amino acids are now in the appropriate place, i.e., the interior and this hydrophobic interaction provides the stability for the subunits to interact in an aqueous environment. 21. In I-cell disease, a patient is missing the enzyme GluNAc phosphotransferase that adds GluNAc-P to the 6 position of a mannose or the enzyme GluNAc phosphoglycosidase that removes the GluNAc leaving the man-6-p. You have a patient whose cells have phenotype similar to those of I-cell disease, lysosomes filled with large inclusions and lysosomal enzymes are secreted into the medium. However, both GluNAc phosphotransferase and GluNAc phosphoglycosidase are normal. A. Propose an hypothesis a molecular basis for this patient's disease (4 pts) i. The cells are missing the Man-6-P receptor ii. The lysosomal enzyme is mutated so it does not get the core-oligosaccharide (this is unlikely since a number of enzymes would have to have been mutated but give credit for this anyway. B. Describe an experiment that you could do to test your hypothesis (explain your answer) (4 pts). i. The secreted enzymes will be properly labeled with Man-6-P and when put into the medium of normal cells will be taken up and delivered to the lysosomes. Add wt lysosomal enzymes to these cells. See if the disease is cured-- If the hypothesis is correct, the mutant cells will not "cured" by addition of wt enzymes. ii. There would be no N-linked oligosaccharide. The cells could be "cured" by addition of wild type lysosomal enzymes. 6

7 22. While working in a lab, you discover a new protein. About 80% of this protein is on the cytosolic surface of transport vesicles and ~ 20% is free in the cytoplasm. By examining its sequence, you find a GTP binding sequence as well as asparagine (N) at position 133 which by analogy to other GTPases is necessary for hydrolysis of GTP to GDP. At its carboxy terminus is a "CAAX" box --a typical isoprenylated proteins. To examine the function of this GTPase, you create mutant forms and express them in cultured cells. When you truncate the protein by removing the CAAX from the the C-terminal, the protein still binds and hydrolyzes GTP but is found only free in the cytoplasm. The cultured cells, which also express the normal (wild type, wt) version of this GTPase, appear normal. Next you change the amino actid at position 133 from N to I. This mutant protein can no longer hydrolyze GTP into GDP. When expressed in the cells, this mutation causes the cells to become filled with small vesicles and eventually to die. When you express a double mutant both missing the CAAX box and with I replacing N at 133, the cells appear normal. A. Do you think your GTPase is required for the formation of vesicles, for vesicle fusion with target membranes, or for both? (2 pts) It is clearly required for vesicle fusion. The experiment does not define whether it is required for vesicle formation since wt GTPase is always present. B. Diagram how you think the wild type GTPase functions. (3 pts) Like a Rab protein, binding to a vesicle upon stimulation to release GDP and bind GTP, the isoprenyl tail is exposed, it inserts into the vesicle membrane, making the protein vesicle associated. Upon interacting with an appropriate target membrane, the GTP is hydrolyzed to GDP, allowing the protein to leave the membrane and allowing the two membranes to fuse. C. What should some of the wild type GTPase be free in the cytosol and some on vesicles? (2 pts) The membrane bound form is the GTP form. The GDP form in the cytoplasm is a recycling pool available for newly formed vesicles. D. What cytosol/membrane distribution do you expect for the I-133 mutant? explain (2 pts) It will be all membrane bound. E. Why might expression of the I-133, in the presence of wildtype protein, cause accumulation of vesicles? (2 pts) The GTP-form of the protein inhibits fusion even when the wt protein has hydrolyzed its GTP and returned to the cytoplasm F. How do you explain the phenotype of the double mutant? (2 pts) In the absence of an isoprenyl tail, the GTP-bound protein can not bind to the membrane. In the cytoplasm it has no effect. 7